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Impedances parallel

  1. Dec 12, 2009 #1
    to find the the total impedance of a circuit i use 1/Zt=1/Z+1/Z...ect. this way is quite lenghty when dealing with more than two impedances.

    i wonder if there is a qicker way of doing it maybe by just using the magnitude of the capacitor to get rid of the imaginary part.

    for example if i had a risistor , a capacitor. and a inductor in series with a risistor.

    what i did was find the impedances of each branch then used the eq ZT=(Z1.Z2)/Z1+Z2 to find the impedance of the first to branch then simpify the circuit replacing the two branches with a single impedance and used the eq again.

    but it doesn't seem to agree with my lecture's answer.
  2. jcsd
  3. Dec 12, 2009 #2


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    It is messier than series circuits.

    I assume some voltage and then work out the currents for the branches of the parallel network.

    Then you can just add the currents to get a total current and then get back to an impedance by dividing the voltage by this current.

    I suppose it amounts to the same process. Because it involves smaller steps, you can track down atithmetic errors easier, I guess.

    If you have capacirors in parallel, you can certainly simplify those (by just adding the capacitances) , but not if they have series resistors or other extra components.
  4. Dec 12, 2009 #3
    What it i was just given a current source and no voltage was given. i guess i would have to do it the other way. i tend to mess up the arithmetic since i havn't covered complex numbers yet. If u have a complex number on the bottom of the first equation then i am not sure what to do to get rid of it.
  5. Dec 12, 2009 #4


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    It is very easy to get the arithmetic wrong with these, so it is important to lay the solution out well so you can check it through afterwards. Explain what you are doing and leave blank lines between calculations.

    If you haven't done complex numbers, you will find these calculations pretty weird.

    You may be able to follow this example:

    complex no calc.PNG

    It starts off with a complex number 12 - j40 in the bottom line. See what happens when you multiply by 12 + j40 / 12 + J40. The J's vanish and you get something you can deal with.
  6. Dec 13, 2009 #5
    Thank you. This should help me.
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